Problem: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $n \neq 0$. $a = \dfrac{7}{28n + 7} \times \dfrac{12n^2 + 3n}{6n} $
Answer: When multiplying fractions, we multiply the numerators and the denominators. $a = \dfrac{ 7 \times (12n^2 + 3n) } { (28n + 7) \times 6n } $ $ a = \dfrac {7 \times 3n(4n + 1)} {6n \times 7(4n + 1)} $ $ a = \dfrac{21n(4n + 1)}{42n(4n + 1)} $ We can cancel the $4n + 1$ so long as $4n + 1 \neq 0$ Therefore $n \neq -\dfrac{1}{4}$ $a = \dfrac{21n \cancel{(4n + 1})}{42n \cancel{(4n + 1)}} = \dfrac{21n}{42n} = \dfrac{1}{2} $